Horizontal acceleration of an object moving down a parabolic track

Question

PROBLEM: The figure above shows an accelerometer, a device for measuring the horizontal acceleration of cars and airplanes. A ball is free to roll on a parabolic track described by the equation $y=x^2$, where both x and y are in meters. A scale along the bottom is used to measure the ball’s horizontal position x. Find an expression that allows you to use a measured position x (in m) to compute the acceleration $a_x$ (in $m/s^2$). (For example, $a_x = 3x$ is a possible expression.)

PROFESSOR'S EXPLANATION: By looking at a small portion of the parabola, the balls motion can be approximated to a ramp, with a base of dx, a height of dy, and an angle of $\theta$. The two forces acting on the ball are $F_g$ and the normal force, n. The normal force can be split into a horizontal component parallel to the base of the ramp, and a vertical component parallel to the height of the ramp. The horizontal component equals $n*sin(\theta)$, and the vertical component equals $n*cos(\theta)$.

The next step is what confuses me. He says that $F_g = n*cos(\theta)$. Thus $mg = n*cos(\theta)$ and $n = \frac{mg}{cos(\theta)}$. This implies the sum of the forces vertically is 0, which is confusing because the ball is accelerating down the ramp. This also conflicts with what I've learned before about motion down a ramp, where the normal force is usually $mg * cos(\theta)$

If you continue with his explanation however, everything else makes sense and you reach the correct answer. By substituting what you got for n into the horizontal component of the normal force, you get $F_{x net} = \frac{mgsin(\theta)}{cos(\theta)} = mgtan(\theta)$. This leads to $ma_x = mgtan(\theta)$ which leads to $a_x = gtan(\theta) = g\frac{dy}{dx}$. Looking at the original problem and our theoretical ramp, $\frac{dy}{dx} = 2x$. Thus $a_x = g2x$.

Why is the normal force in this problem $\frac{mg}{cos(\theta)}$ and not $mg * cos(\theta)$ like it normally is? How can you set $F_g$ equal to $n*cos(\theta)$ when the ball is accelerating down the ramp? Any help is appreciated.

Answer 1

I believe your confusion comes from thinking the ball is accelerating down the ramp. The horizontal acceleration is measured when the ball is in equilibrium with the parabolic block. The ball is actually stationary with respect to the block. Looking at the $m\ a=\text{$\Sigma $F}$ in the y direction:

$m\ a_y=m\ g - n \cos (\theta )$

but in equilibrium $a_y$ is zero and we have your professor's first equation from which we can calculate $n$.

In equilibrium the $x$ acceleration of the ball is the same as the $x$ acceleration of the block and we have

$m\ a_x=n \sin (\theta )$

and substituting for $n$ gives your other equation.

Answer 2

If I understand the question correctly, the figure below may help with some of the geometry. The key idea is that when you have non-zero horizontal acceleration, the ball will move up the curve until it stops at a certain location. Seeing that the ball is in a fixed position relative to the curve/accelerometer and the geometry figure below should allow you to work out the rest.

I hope this helps.

Answer 3

First, let me say that I can't see the images owing to a security issue, but I think I can infer the geometry of the problem. I think you are (or were: so long ago!) right to be confused: the net vertical force is not zero. Either the professor made a mistake or he meant something else. The only additional observation I can offer is this: The only force that is equal to n*cos(theta) is that component of the gravitational force (vertical) that you have to subtract from the total gravitational force "Fg,total" in order to get the net vertical force acting on the ball.